Akshat Y. answered • 06/17/21

Calculus Tutor with 5 on AP Exam and 3 Years Tutoring Experience

Hi Stormee,

We can write the equation of the tangent line in point-slope form, which is y - y_{0} = m(x - x_{0}), where (x_{0}, y_{0}) is the point given to us and m is the slope (or derivative dy/dx) at this point. We are given the point (-2, 3), so now we have y - 3 = m(x + 2).

To find the derivative at this point, we first have to find dy/dx of this function using implicit differentiation with respect to x. The work is shown below.

d/dx [ln(10 - y^{2}) + 16] = d/dx [-2x^{3}]

⇒ d/dx[10 - y^{2}] * 1 / (10 - y^{2}) = -6x^{2} (the left-hand side is the chain rule used on ln. Remember that d/dx [ln(x)] = 1/x.

⇒ -2y * dy/dx * 1 / (10 - y^{2}) = -6x^{2} (The resulting dy/dx is also a result of the chain rule).

Solving the above equation for the slope dy/dx, we get

dy/dx = 3x^{2}(10 - y^{2}) / y.

To find the derivative at the given point (-2, 3), we plug in x = -2 and y = 3 into dy/dx, getting

dy/dx at (-2, 3) = m = 3(-2)^{2}(10 - 3^{2}) / 3 = 4

Substituting this into the equation for the line, we have our answer of

**y - 3 = 4(x + 2)**

or in slope-intercept form,

**y = 4x + 11**